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| Tags: accurately, believe, correct, describes, essentially, general, given, math, nature, relativity, results |
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#21
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Tom Roberts wrote:
Joe Fischer wrote: On Tue, 29 Nov 2005 05:27:50 GMT, Tom Roberts wrote: Joe Fischer wrote: On Mon, 28 Nov 2005 15:27:16 -0600, Tom Roberts wrote: Hmmm. In GR, gravitation is not "additive", if by that term you really mean linear. No, I don't mean linear, I mean additive, [...] Then you need to define what you mean by "additive". http://www.faqs.org/faqs/astronomy/f...section-5.html The only appearance of "additive" on that page is used in the sense of "linear". The only use of 'additive' on that page is the following: Second, gravity is additive. Planets and stars are very nearly electrically neutral, so the forces exerted by positive and negative charges tend to cancel out. The are not using 'additive' in any sort of technical sense. The apparent meaning is that because gravity is always an attractive force, as opposed to electrical forces which can be attractive or repulsive, gravitational forces "add up" to large forces. Electrical forces are not "additive" because they tend to cancel out. Of course for the mathematically minded there is nothing non-"additive" about the fact that adding up a bunch of positive and negative numbers may result in 0. Stephen |
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#22
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On Tue, 29 Nov 2005 10:37:52 -0600, Tom Roberts wrote:
Joe Fischer wrote: On Tue, 29 Nov 2005 05:27:50 GMT, Tom Roberts wrote: Joe Fischer wrote: On Mon, 28 Nov 2005 15:27:16 -0600, Tom Roberts wrote: Hmmm. In GR, gravitation is not "additive", if by that term you really mean linear. No, I don't mean linear, I mean additive, [...] Then you need to define what you mean by "additive". http://www.faqs.org/faqs/astronomy/f...section-5.html The only appearance of "additive" on that page is used in the sense of "linear". Tom Roberts I don't see how you can ascertain that, the sentence is rather short, " Second, gravity is additive.". I don't see why "additive" should mean linear, the author of that FAQ surely knew how to spell linear. Additive simply means the more matter and energy there is, the more gravity. It doesn't seem to imply linear or any parameters except more = more. In the Divergent Matter model, it is obvious gravity is additive, because only with contact is there a "force" involved, and that force is an inertial resistance to the expansion of matter. Because of this and other factors, I believe it is wrong to say that fusion "releases" energy, I believe it is squeezed out, and this should mean no comtainment is possible except inertial. The Divergent Matter model obviously rules out shielding or artificial gravity using fields, too bad for the dreamers. It also slightly changes the dynamics in the sun, if instead of an attractive "field", it is the inertial resistance of the upper layers that resist the expansion, then the pressures should be different at various depths. It should also make black holes less likely, without the concept of "attraction", a singularity is iimpossible, inertial resistance to expansion will compress, but with some limits. For a given radius sphere, adding more mass or energy, does add to the gravity, because the expansion increases. If gravity were an attractive field it would seem possible to create a gravity beam where the strength or flux could be greater above the surface, but this is impossible in Divergent Matter, and impossible in nature, gravity always falls off above the surface, but may fall off or even increase below the surface depending on the distribution of mass and energy. But the upeard acceleration of the surface is not the primary factor in the Divergent Matter model, the expansion velocity of the surface represents a considerable amount of energy, and produces most of the geometry changes. I will try to find a better description of additive gravity, or try to write one. Joe Fischer |
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#23
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Joe Fischer wrote:
On Tue, 29 Nov 2005 10:37:52 -0600, Tom Roberts wrote: Joe Fischer wrote: On Tue, 29 Nov 2005 05:27:50 GMT, Tom Roberts wrote: Joe Fischer wrote: On Mon, 28 Nov 2005 15:27:16 -0600, Tom Roberts wrote: Hmmm. In GR, gravitation is not "additive", if by that term you really mean linear. No, I don't mean linear, I mean additive, [...] Then you need to define what you mean by "additive". http://www.faqs.org/faqs/astronomy/f...section-5.html The only appearance of "additive" on that page is used in the sense of "linear". Tom Roberts I don't see how you can ascertain that, the sentence is rather short, " Second, gravity is additive.". I don't see why "additive" should mean linear, the author of that FAQ surely knew how to spell linear. Additive simply means the more matter and energy there is, the more gravity. It doesn't seem to imply linear or any parameters except more = more. In the Divergent Matter model, it is obvious gravity is additive, because only with contact is there a "force" involved, and that force is an inertial resistance to the expansion of matter. Because of this and other factors, I believe it is wrong to say that fusion "releases" energy, I believe it is squeezed out, and this should mean no comtainment is possible except inertial. The Divergent Matter model obviously rules out shielding or artificial gravity using fields, too bad for the dreamers. It also slightly changes the dynamics in the sun, if instead of an attractive "field", it is the inertial resistance of the upper layers that resist the expansion, then the pressures should be different at various depths. It should also make black holes less likely, without the concept of "attraction", a singularity is iimpossible, inertial resistance to expansion will compress, but with some limits. For a given radius sphere, adding more mass or energy, does add to the gravity, because the expansion increases. If gravity were an attractive field it would seem possible to create a gravity beam where the strength or flux could be greater above the surface, but this is impossible in Divergent Matter, and impossible in nature, gravity always falls off above the surface, but may fall off or even increase below the surface depending on the distribution of mass and energy. But the upeard acceleration of the surface is not the primary factor in the Divergent Matter model, the expansion velocity of the surface represents a considerable amount of energy, and produces most of the geometry changes. I will try to find a better description of additive gravity, or try to write one. Joe Fischer You're good.!! Squeeze up the margins a little. brian a m stuckless |
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#24
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Koobee Wublee:
Dr. Roberts wrote too confidently: Hmmm. That basically comes from writing down the Lagrangians for all known forces and examining their coupling constants. And do these Lagrangians follow the contraints put forth by the Lagrangian Methods? By definition, the lagrangian satisfies the euler-lagrange equations. In general, that isn't particularly useful, although it might be more useful in general relativity than in high energy physics. Can you [plural] always identify an action to be minimzed and the parameter to be minimized? Of course not. There is no formula to derive a lagrangian. You use some intuition and make an intelligent guess. If there was a recipe, we would have a theory of everything, which is guaranteed to be right. Since no one knows what the correct physics is, one must construct a lagrangian that reproduces the known physics. In doing so, it's almost a guarantee that you will be stuck with a prediction of some unkown physics, since there are almost certainly different lagrangians which can reproduce the data which is not controversial. [...] Dr. Roberts wrote: The one for Gravity is smaller than the others by an enormous factor. For instance, between two electrons at a given distance the ratio of gravitational force to electrostatic force is about 10^-42 -- that is INCREDIBLY TINY. What are the Lagrangians you [plural] have identified for electrosatic and magnetic forces? That is simple: L = (s+)[p/ - m] s - j^uA_u + (1/4)F^uv F_uv p/ = i\gamma^u d_u, j^u = i(s+)\gamma^v (s) \gamma^u are dirac matrices and s,s+ are a spinor and conjugate spinor. [...] So, Newton established a thoery on how two objects with mass attracts each other. Actually he didn't. What he did was just state the geometric fall off, assuming a 3-d euclidean space. It's gauss' law with two masses and G instead of two charges. He humbly pointed that the exact mechanism on how this attractive force comes about still eludes mankind. Grossmann/Einstein/Hilbert came up with a theory explaining how two objects would behave with observed attractive force through the curvature of spacetime suround massive objects. Just because Einstein did not humbly like Newton did pointed out exactly why mass and energy would manifest a curvature in spacetime, there is no closer closure to the understanding on how two objects with mass are going to attract each other. The reason einstein didn't give such an explanation, is because einstein didn't start off assuming the geomtery was some particular way in the first place. The reason that gravity is a force in newtonian mechanincs is because newton made an assumption about space and time and his assumption required gravity to be a force. If you prefer spacetime to be flat, then the burden is on you to explain why flat spacetime causes a force of gravity. In curved spacetime, there is no gravitational force and most everyone believes that you can't create or eliminate a real force by changing coordinates. You are free to believe otherwise, however. [...] Even through the curvature of spacetime, one can also write an equation describing the observed energy or the observed mass of an object trapped in this curvature of spacetime. An object which ``trapped'' is only trapped because the future of every timelike geodesic is inside the region in which the object is trapped. What you are trying to do is equivalent to asking why a meridian is on the earth's surface is trapped to lie on the surface of the earth. The reason is, because the geodesics are great circles by definition. It makes no sense to ask why none of them encircle the moon. If they did, they wouldn't be geodesics on the earth. [...] Please show us your Lagrangian(s) involved. What coupling constants are you referring to? Calculate the force produced by two 1 kg spherers of protons separated by 100,000 km. That's about 6 x 10^26 protons, or about the same number of protons as found in a little over a gallon of water. Since there is 1 x 10^-19 coul/proton, that's about 95 million coulombs per sphere. I get about 800,000,000 newtons. If that failed to impress you, calculate the classical acceleration on the electron orbiting at the bohr radius with a velocity of c/137. If that still failed to be persuasive, calculate the net charge imbalance the earth could tolerate without being blown apart. Next, note that the strong nuclear force is over 1000 times stronger. Just using the most naive estimate of about 8 MeV for the kinetic energy of a generic nucleon in the nucleus, figure out the force which holds a neutron within about 1.5 fm of a nuclues (at the most). |
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#25
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"Bilge" wrote in message By definition, the lagrangian satisfies the euler-lagrange equations. In general, that isn't particularly useful, although it might be more useful in general relativity than in high energy physics. Lagrangian is defined as the density to an action. It is a very abstract measurement. Another one comes to mind is the entropy. Since not all actions can be minimized, Euler-Lagrange equations do not in general come about. Can you [plural] always identify an action to be minimzed and the parameter to be minimized? Of course not. There is no formula to derive a lagrangian. You use some intuition and make an intelligent guess. If there was a recipe, we would have a theory of everything, which is guaranteed to be right. Since no one knows what the correct physics is, one must construct a lagrangian that reproduces the known physics. In doing so, it's almost a guarantee that you will be stuck with a prediction of some unkown physics, since there are almost certainly different lagrangians which can reproduce the data which is not controversial. You cannot define a term and call that a Lagrangian. A Langrangian has to be the density of an action. Thus, you have to qualify the term you have created according to the conditions of the Lagrangian Method in order to call it a Lagrangian. What are the Lagrangians you [plural] have identified for electrosatic and magnetic forces? That is simple: L = (s+)[p/ - m] s - j^uA_u + (1/4)F^uv F_uv p/ = i\gamma^u d_u, j^u = i(s+)\gamma^v (s) \gamma^u are dirac matrices and s,s+ are a spinor and conjugate spinor. How is this Lagrangian a desnity of an action? What action is it? Is there a minimum to this action? [...] So, Newton established a thoery on how two objects with mass attracts each other. Actually he didn't. What he did was just state the geometric fall off, assuming a 3-d euclidean space. It's gauss' law with two masses and G instead of two charges. What you are saying does not contradict what I said above. He humbly pointed that the exact mechanism on how this attractive force comes about still eludes mankind. Grossmann/Einstein/Hilbert came up with a theory explaining how two objects would behave with observed attractive force through the curvature of spacetime suround massive objects. Just because Einstein did not humbly like Newton did pointed out exactly why mass and energy would manifest a curvature in spacetime, there is no closer closure to the understanding on how two objects with mass are going to attract each other. The reason einstein didn't give such an explanation, is because einstein didn't start off assuming the geomtery was some particular way in the first place. The reason that gravity is a force in newtonian mechanincs is because newton made an assumption about space and time and his assumption required gravity to be a force. What assumptiom did Newton make about space and time? If you prefer spacetime to be flat, then the burden is on you to explain why flat spacetime causes a force of gravity. In curved spacetime, there is no gravitational force and most everyone believes that you can't create or eliminate a real force by changing coordinates. You are free to believe otherwise, however. This is easy. Even GR explains a gravitational force through the curvature of spacetime. For example, in the environment we are familiar with (spacetime with Schwarzschild metric), the following equation can be derived by minimizing x^0 (observed time, t). E = m c^2 sqrt(1 - 2 U) / sqrt(1 - B^2) Where E is the observed energy of an object moving and trapped in the curvature of spacetime and ** U = G M / c^2 / r ** B^2 c^2 = (dr/dt)^2 / (1 - 2 U) + r^2 (dH/dt)^2 / (1 - 2 U)^2 +... ** All other parameters self explanatory through content of usage Since energy is constant, you can find the exact motion of this object from the equation above. In there, you will find your gravitational force according to GR. In Newtonian and thus classical sense, ** 1 2 U ** 1 B^2 Thus, E ~= m c^2 (1 - U + B^2 / 2) Where ** m B^2 c^2 / 2 = kinetic energy ** m U c^2 = - potential energy [...] Even through the curvature of spacetime, one can also write an equation describing the observed energy or the observed mass of an object trapped in this curvature of spacetime. An object which ``trapped'' is only trapped because the future of every timelike geodesic is inside the region in which the object is trapped. What you are trying to do is equivalent to asking why a meridian is on the earth's surface is trapped to lie on the surface of the earth. The reason is, because the geodesics are great circles by definition. It makes no sense to ask why none of them encircle the moon. If they did, they wouldn't be geodesics on the earth. Please allow me to rephrase. Instead of 'strapped', a better word is 'immersed'. [...] Please show us your Lagrangian(s) involved. What coupling constants are you referring to? Calculate the force produced by two 1 kg spherers of protons separated by 100,000 km. That's about 6 x 10^26 protons, or about the same number of protons as found in a little over a gallon of water. Since there is 1 x 10^-19 coul/proton, that's about 95 million coulombs per sphere. I get about 800,000,000 newtons. If that failed to impress you, calculate the classical acceleration on the electron orbiting at the bohr radius with a velocity of c/137. If that still failed to be persuasive, calculate the net charge imbalance the earth could tolerate without being blown apart. Next, note that the strong nuclear force is over 1000 times stronger. Just using the most naive estimate of about 8 MeV for the kinetic energy of a generic nucleon in the nucleus, figure out the force which holds a neutron within about 1.5 fm of a nuclues (at the most). You still have not shown me the Lagrangian involved. Be prepared. If you can show me a Lagrangian, I am going to ask you what action is this a density of. |
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#26
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"Koobee Wublee" wrote in message news:Rx9kf.632$La5.338@fed1read01... "Bilge" wrote in message "You use some intuition and make an intelligent guess. If there was a recipe, we would have a theory of everything, which is guaranteed to be right." "Intuition"... "guess"... Is it remotely intelligent to make such a suggestion? One may as well go back to geocentric universe of Ptolemy, it was "intuitive". How does a ****in' idiot like Bilge ever survive ordinary daily life? Does anyone want to make an intelligent guess where Bilge wants replies redirected to? Intuition tells me it's alt.morons, his favourite stomping ground. He feels at home there. Is it not intuitively obvious that one cannot have an intelligent, reasoned discussion with intuitive guesses, intelligent or otherwise? Be reasonable! See it Bilge's way. If you don't, be redirected to alt.morons. Androcles. |
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#27
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Koobee Wublee:
"Bilge" wrote in message By definition, the lagrangian satisfies the euler-lagrange equations. In general, that isn't particularly useful, although it might be more useful in general relativity than in high energy physics. Lagrangian is defined as the density to an action. It is a very abstract Look, if you don't want to believe me, suit yourself and go on trying to disprove whatever you find time to misconstrue. [...] That is simple: L = (s+)[p/ - m] s - j^uA_u + (1/4)F^uv F_uv p/ = i\gamma^u d_u, j^u = i(s+)\gamma^v (s) \gamma^u are dirac matrices and s,s+ are a spinor and conjugate spinor. How is this Lagrangian a desnity of an action? What action is it? Is there a minimum to this action? Since you call yourself an expert, it ought to be obvious - especially since I told you what all of the pieces were not to mention being the most famous lagrangian in history as well as being the most precisely tested theory in science. If you want to be an expert - be an expert. [...] If you prefer spacetime to be flat, then the burden is on you to explain why flat spacetime causes a force of gravity. In curved spacetime, there is no gravitational force and most everyone believes that you can't create or eliminate a real force by changing coordinates. You are free to believe otherwise, however. This is easy. Even GR explains a gravitational force through the curvature of spacetime. Gravity is not a force in general relativity. If that hasn't sunk in, you need to go back and get a bit more expertise. |
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